A heavy box is to dragged along a rough horizontal floor. To do so, person A pushes it at an angle 30^o from the horizontal

Q: A heavy box is to dragged along a rough horizontal floor. To do so, person $A$ pushes it at an angle $30^o$ from the horizontal and requires a minimum force $F_A$, while person $B$ pulls the box at an angle $60^o$ from the horizontal and needs minimum force $F_B$. If the coefficient of friction between the box and the floor is $\frac{{\sqrt 3 }}{5}$ , the ratio $\frac{{{F_A}}}{{{F_B}}}$ is

d $\begin{array}{l} {F_A} = \frac{{\mu mg}}{{\sin \theta - \mu \cos \theta }}\\ similarly,\\ {F_B} = \frac{{\mu mg}}{{\sin \theta + \mu \cos \theta }}\\ \,\therefore \,\frac{{{F_A}}}{{{F_B}}} = \frac{{\frac{{\mu mg}}{{\sin \theta - \mu \cos \theta }}}}{{\frac{{\mu mg}}{{\sin \theta + \mu \cos \theta }}}}\,\,\,\,\,\\ \, = \frac{{\sin {{60}^ \circ } - \frac{\begin{array}{l} \mu mg\\ \sqrt 3 \end{array}}{5}\cos \,{{60}^ \circ }}}{{\sin {{30}^ \circ } + \frac{\begin{array}{l} \mu mg\\ \sqrt 3 \end{array}}{5}\cos \,{{30}^ \circ }}}\,\, \end{array}$ $\begin{array}{l} \,\,\left[ {\mu = \frac{{\sqrt 3 }}{5}\,given} \right]\,\\ = \frac{{\sin \,{{30}^{ \circ \,}} + \frac{{\sqrt 3 }}{5} - \cos \,{{30}^ \circ }}}{{\sin \,{{60}^ \circ } - \frac{{\sqrt 3 }}{5}\cos \,{{60}^ \circ }}}\\ = \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{5} \times \frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{5} \times \frac{1}{2}}} \end{array}$ $\begin{array}{l} = \frac{{\frac{1}{2}\left( {1 + \frac{3}{5}} \right)}}{{\frac{{\sqrt 3 }}{5}\left( {1 - \frac{1}{5}} \right)}} = \frac{{\frac{1}{2} \times \frac{8}{5}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}}\\ = \frac{{\frac{8}{{10}}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}} = \frac{8}{{\sqrt 3 \times 4}} = \frac{2}{{\sqrt 3 }} \end{array}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A heavy box is to dragged along a rough horizontal floor. To do so, person $A$ pushes it at an angle $30^o$ from the horizontal and requires a minimum force $F_A$, while person $B$ pulls the box at an angle $60^o$ from the horizontal and needs minimum force $F_B$. If the coefficient of friction between the box and the floor is $\frac{{\sqrt 3 }}{5}$ , the ratio $\frac{{{F_A}}}{{{F_B}}}$ is

A$\sqrt 3 $
B$\frac{5}{{\sqrt 3 }}$
C$\sqrt {\frac{3}{2}} $
D$\frac{2}{{\sqrt 3 }}$

JEE MAIN 2014, Diffcult

STD 11 Science
NEET
STD 11 - 4.2. friction
Physics

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